Quadratic Inequalities

A quadratic inequality is an inequality that contains a quadratic expression. The standard forms are $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$, $ax^2 + bx + c \geq 0$, or $ax^2 + bx + c \leq 0$.

Solving a quadratic inequality means finding all the values of $x$ that make the inequality true. Instead of a single number, the solution is usually a range (or interval) of numbers on the number line.

How to Solve Quadratic Inequalities

To solve a quadratic inequality, follow these three main steps:

1. Move everything to one side: Ensure one side of the inequality is zero.
2. Find the critical points: Treat the inequality symbol as an equal sign and solve the quadratic equation (usually by factoring) to find the roots. These roots divide the number line into distinct intervals.
3. Determine the correct intervals: Use a sign chart to test a number in each interval, or quickly sketch the parabola to see where the graph is above or below the x-axis.

Example 1: Solving with a Sign Chart

Solve: $x^2 - 5x + 6 > 0$

Step 1: Find the critical points. Set the expression to zero and factor: $x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0$ The critical points are $x = 2$ and $x = 3$.

Step 2: Set up intervals. These points divide the number line into three intervals:

• $x < 2$
• $2 < x < 3$
• $x > 3$

Step 3: Test each interval. Pick a test value in each interval to see if $(x - 2)(x - 3)$ is positive or negative:

• For $x = 0$ (in $x < 2$): $(0 - 2)(0 - 3) = (-2)(-3) = 6$. Positive!
• For $x = 2.5$ (in $2 < x < 3$): $(2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25$. Negative.
• For $x = 4$ (in $x > 3$): $(4 - 2)(4 - 3) = (2)(1) = 2$. Positive!

Since our original inequality is $> 0$ (we want the positive values), the solution is the intervals where the result is positive.

Solution: $x < 2$ or $x > 3$

Example 2: Solving by Sketching the Parabola

Solve: $-x^2 + 4x - 3 \geq 0$

Step 1: Make the leading coefficient positive (optional but helpful). Multiply the entire inequality by $-1$. Remember that multiplying or dividing an inequality by a negative number flips the inequality sign: $x^2 - 4x + 3 \leq 0$

Step 2: Find the critical points. Factor the quadratic: $(x - 1)(x - 3) = 0$ The critical points are $x = 1$ and $x = 3$.

Step 3: Sketch the parabola. Look at the graph of $y = x^2 - 4x + 3$. Since the $x^2$ coefficient is positive ($1 > 0$), the parabola opens upwards (like a "U" shape). It crosses the x-axis at $x = 1$ and $x = 3$.

We want to know where the parabola is $\leq 0$ (below or touching the x-axis). Looking at our mental "U" shape, the graph dips below the x-axis strictly between the roots 1 and 3.

Solution: $1 \leq x \leq 3$